Answer:
Refer below.
Explanation:
VPC, Subnets, Route Table(s), Nat Gateway, and Internet Gateway. These are the least required services to provide internet access to a private EC2 instance. NAT gateway requires an internet gateway.
Answer:
B. R is NP Hard
Explanation:
Given:
S is an NP complete problem
Q is not known to be in NP
R is not known to be in NP
Q is polynomial times reducible to S
S is polynomial times reducible to R
Solution:
NP complete problem has to be in both NP and NP-hard. A problem is NP hard if all problems in NP are polynomial time reducible to it.
Option B is correct because as given in the question S is an NP complete problem and S is polynomial times reducible to R.
Option A is not correct because R is not known to be in NP
Option C is not correct because Q is also not known to be in NP
Option D is not correct because Q because no NP-complete problem is polynomial time reducible to Q.
Answer:
is there a picture you can attach to show the cell, row, and column?
Explanation:
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
4. B
for # 5 what illustration are u talking about
6. D
7. D
8. don't understand that question
sorry these questions might be wrong